#测试点1、6无法通过
题目:
以下是代码:
#include<stdio.h>
int main()
{
int N;
scanf("%d",&N);
int s[N];
for(int i=0;i<N;i++)
scanf("%d",&s[i]);
int a0[1000],a1[1000],a2[1000],a3[1000],a4[1000];
int j=0,k=0,m=0,n=0,p=0;
for(int i=0;i<N;i++)
{
if(s[i]%5==0)
{
a0[j]=s[i];
j++;
}
else if(s[i]%5==1)
{
a1[k]=s[i];
k++;
}
else if(s[i]%5==2)
{
a2[m]=s[i];
m++;
}
else if(s[i]%5==3)
{
a3[n]=s[i];
n++;
}
else
{
a4[p]=s[i];
p++;
}
}
//余数为0:
if(j==0)
printf("N ");
else
{
int sum=0;
for(int i=0;i<j;i++)
{
if(a0[i]%2==0)
sum+=a0[i];
}
printf("%d ",sum);
}
//余数为1
if(k==0)
printf("N ");
else
{
int sum=0;
for(int i=0;i<k;i++)
{
if(i%2==0)
sum=sum+a1[i];
else
sum=sum-a1[i];
}
printf("%d ",sum);
}
//余数为2
if(m==0)
printf("N ");
else
printf("%d ",m);
//余数为3
if(n==0)
printf("N ");
else
{
double average=0;
for(int i=0;i<n;i++)
average+=a3[i];
average=1.0*average/n;
printf("%.1f ",average); // %.nf表示保留n位小数
}
//余数为4
if(p==0)
printf("N");
else
{
int max=a4[0];
for(int i=0;i<p;i++)
{
if(max<a4[i])
max=a4[i];
}
printf("%d",max);
}
return 0;
}
请求点拨
你这个代码的算话实现的结构有些问题:
我给你我的代码吧:
#include <stdio.h>
int main() {
int N;
scanf("%d", &N);
int arr[N], A[6] = {0}, cntA[6] = {0};
for (int i = 0; i < N; i++) {
scanf("%d", &arr[i]);
if (arr[i] % 10 == 0) { //能被5整除的偶数
A[1] += arr[i];
cntA[1]++;
} else if (arr[i] % 5 == 1) { //被5除后余1
cntA[2]++;
if (cntA[2] % 2 == 1) { //第奇数个
A[2] += arr[i];
} else {
A[2] -= arr[i];
}
} else if (arr[i] % 5 == 2) { //被5除后余2
A[3]++;
cntA[3]++;
} else if (arr[i] % 5 == 3) { //被5除后余3
A[4] += arr[i];
cntA[4]++;
} else if (arr[i] % 5 == 4 && A[5] < arr[i]) { //被5除后余4
A[5] = arr[i];
cntA[5]++;
}
}
for (int i = 1; i <= 5; i++) {
if (cntA[i] == 0) {
if (i != 1)
printf(" ");
printf("N");
} else if (i == 1) {
printf("%d", A[1]);
} else if (i == 2 || i == 3 || i == 5) {
printf(" %d", A[i]);
} else { // i = 4
printf(" %.1f", (double)A[i] / cntA[4]);
}
}
printf("\n");
return 0;
}
您好,我是有问必答小助手,您的问题已经有小伙伴帮您解答,感谢您对有问必答的支持与关注!数字分类?
// PAT BasicLevel 1012
// https://pintia.cn/problem-sets/994805260223102976/problems/994805311146147840
#include <iostream>
using namespace std;
int main()
{
// 数字个数
int n;
cin >> n;
// 获取数字
int* numArr=new int[n];
for(int i=0;i<n;++i){
cin >> numArr[i];
}
// 遍历数组,计算A1至A5
int a1=0,a1Count=0;
int a2=0,flag=1,a2Count=0;
int a3=0,a3Count=0;
double a4Sum=0,a4Count=0;
int a5=0,a5Count=0;
for(int i=0;i<n;++i){
switch(numArr[i]%5){
case 0:
if(numArr[i]%2==0){
a1+=numArr[i];
a1Count++;
}
break;
case 1:
a2+=flag*numArr[i];
flag=-flag;
a2Count++;
break;
case 2:
a3++;
a3Count++;
break;
case 3:
a4Sum+=numArr[i];
a4Count++;
break;
case 4:
if(numArr[i]>a5){
a5=numArr[i];
a5Count++;
}
break;
}
}
// 输出A1至A5
if(a1Count>0){
cout << a1 << ' ';
}else{
cout <<"N ";
}
if(a2Count>0){
cout << a2 << ' ';
}else{
cout << "N ";
}
if(a3Count>0){
cout << a3 << ' ';
}else{
cout << "N ";
}
if(a4Count>0){
printf("%.1lf ", a4Sum / a4Count);
}else{
cout << "N ";
}
if (a5Count > 0){
cout << a5;
}
else{
cout << 'N';
}
delete[] numArr;
//system("pause");
return 0;
}
用switch case